Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Derivatives of Polar Form

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Derivatives Of Polar Form

For the polar equation $r = cos(\theta) + sin(\theta),$ $0<\theta<2\pi$ , find $\frac{dy}{dx}$ when $\theta = \pi/3$ .

Possible Answers:

$\sqrt3-2$

$1+\sqrt3$

$\sqrt2/2$

None of the other answers.

Correct answer:

$\sqrt3-2$

Explanation:

When
$\theta = \pi/3, r = cos(\pi/3)+sin (\pi/3) = \frac{1+\sqrt3}{2}$ .

Taking the derivative of our given equation with respect to $\theta$ , we get

$\frac{dr}{d\theta} = -sin(\theta)+cos(\theta)$

To find $\frac{dy}{dx}$ , we use

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{dr}{d\theta}sin(\theta)+rcos(\theta)}{\frac{dr}{d\theta}cos(\theta)-rsin(\theta)}.$

Substituting our values of $r, \theta$ into this equation and simplifying carefully using algebra, we get the answer of $\sqrt{3}-2$ .

Report an Error

Example Question #1 : Derivatives Of Polar Form

Find the derivative of the following polar equation:

$r=3+8sin\theta$

Possible Answers:

$\frac{8cos\theta sin\theta-6cos\theta }{4cos^2\theta+6sin\theta+3sin^2\theta}$

$\frac{16sin^2\theta+3cos\theta }{8cos^2\theta-3cos\theta-8sin^2\theta}$

$\frac{8cos\theta sin^2\theta+3cos^2\theta }{4cos^2\theta+3sin\theta-6sin^2\theta}$

$\frac{8cos\theta sin\theta+3sin\theta }{4sin^2\theta-3sin\theta-8cos^2\theta}$

$\frac{16cos\theta sin\theta+3cos\theta }{8cos^2\theta-3sin\theta-8sin^2\theta}$

Correct answer:

$\frac{16cos\theta sin\theta+3cos\theta }{8cos^2\theta-3sin\theta-8sin^2\theta}$

Explanation:

Our first step in finding the derivative dy/dx of the polar equation is to find the derivative of r with respect to $\theta$ . This gives us:

$\frac{dr}{d\theta}=8cos\theta$

Now that we know dr/d $\theta$ , we can plug this value into the equation for the derivative of an expression in polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta }{\frac{dr}{d\theta }cos\theta-rsin\theta }=\frac{8cos\theta sin\theta+(3+8sin\theta )cos\theta}{8cos^2\theta-(3+8sin\theta)sin\theta }$

Simplifying the equation, we get our final answer for the derivative of r:

$\frac{dy}{dx}=\frac{16cos\theta sin\theta +3cos\theta }{8cos^2\theta -3sin\theta -8sin^2\theta }$

Report an Error

Example Question #1 : Derivatives Of Polar Form

Evaluate the area given the polar curve: $r=\frac{\theta}{2}$ from $[0,2\pi]$ .

Possible Answers:

$\frac{\pi^3}{3}$

$8\pi$

$\frac{\pi^3}{4}$

$\pi$

$\frac{\pi^3}{6}$

Correct answer:

$\frac{\pi^3}{3}$

Explanation:

Write the formula to find the area in between two polar equations.

$A=\frac{1}{2}\int_{a}^{b}[(\textup{Outer Radius}^2)-(\textup{Inner Radius})^2]\: d\theta$

The outer radius is $r=\frac{\theta}{2}$ .

The inner radius is r=0 .

Substitute the givens and evaluate the integral.

$A=\frac{1}{2}\int_{0}^{2\pi}\left(\frac{\theta}{2}\right)^2\: d\theta=\frac{1}{2}\int_{0}^{2\pi}\frac{\theta^2}{4}\: d\theta=\frac{1}{8}\left[\frac{\theta^3}{3}\right]_{0}^{2\pi}$

$\frac{1}{8}\left[\frac{\theta^3}{3}\right]_{0}^{2\pi}= \frac{1}{8}\left[\frac{(2\pi)^3}{3}\right] = \frac{\pi^3}{3}$

Report an Error

Example Question #2 : Derivatives Of Polar Form

Find the derivative $\frac{dy}{dx}$ of the polar function $r(\theta)=3-4sin(\theta)$ .

Possible Answers:

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

$\frac{dy}{dx}=-4cos(\theta)$

$\frac{dy}{dx}=4cos(\theta)$

$\frac{dy}{dx}=\frac{-4cos(\theta)cos(\theta)+(3-4sin(\theta))sin(\theta)}{-4cos(\theta)sin(\theta)-(3-4sin(\theta))cos(\theta)}$

Correct answer:

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

Explanation:

The derivative of a polar function is found using the formula

$\frac{dy}{dx}=\frac{r'(\theta)sin(\theta)+r(\theta)cos(\theta)}{r'(\theta)cos(\theta)-r(\theta)sin(\theta)}$

The only unknown piece is $r'(\theta)$ . Recall that the derivative of a constant is zero, and that

$\frac{d}{d\theta}sin(\theta)=cos(\theta)$ , so

$r'(\theta)=-4cos(\theta)$

Substiting $r'(\theta)\ and\ r(\theta)$ this into the derivative formula, we find

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos(\theta)cos(\theta)-(3-4sin(\theta))sin(\theta)}$

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

Report an Error

Example Question #2 : Derivatives Of Polar Form

Find the first derivative of the polar function

$r(\theta)=sin(3\theta)$ .

Possible Answers:

$\frac{dy}{dx}=\frac{3cos(3\theta)sin(\theta)+sin(3\theta)cos(\theta)}{3cos(3\theta)cos(\theta)-sin(3\theta)sin(\theta)}$

$\frac{dy}{dx}=\frac{3sin(3\theta)cos(\theta)+cos(3\theta)sin(\theta)}{3sin(3\theta)sin(\theta)-cos(3\theta)cos(\theta)}$

$\frac{dy}{dx}=\frac{sin(3\theta)sin(\theta)+3cos(3\theta)cos(\theta)}{sin(3\theta)cos(\theta)-3cos(3\theta)sin(\theta)}$

$\frac{dy}{dx}=3cos(3\theta)$

Correct answer:

$\frac{dy}{dx}=\frac{3cos(3\theta)sin(\theta)+sin(3\theta)cos(\theta)}{3cos(3\theta)cos(\theta)-sin(3\theta)sin(\theta)}$

Explanation:

In general, the dervative of a function in polar coordinates can be written as

$\frac{dy}{dx}=\frac{r'(\theta)sin(\theta)+r(\theta)cos(\theta)}{r'(\theta)cos(\theta)-r(\theta)sin(\theta)}$ .

Therefore, we need to find $r'(\theta)$ , and then substitute $r(\theta)\ and\ r'(\theta)$ into the derivative formula.

To find $r'(\theta)$ , the chain rule,

$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$ , is necessary.

We also need to know that

$\frac{d}{dx}sin(x)=cos(x)$ .

Therefore,

$r'(\theta)=3cos(3\theta)$ .

Substituting $r(\theta)\ and\ r'(\theta)$ into the derivative formula yields

$\frac{dy}{dx}=\frac{3cos(3\theta)sin(\theta)+sin(3\theta)cos(\theta)}{3cos(3\theta)cos(\theta)-sin(3\theta)sin(\theta)}$

Report an Error

Example Question #3 : Derivatives Of Polar Form

Find the derivative of the following function:

$r=2\cos(\theta)-1$

Possible Answers:

$\frac{-2sin^2(\theta)+2\cos^2(\theta)-\cos(\theta)}{4\cos(\theta)\sin(\theta)-\sin(\theta)}$

$\frac{-2sin^2(\theta)+2\cos^2(\theta)-\cos(\theta)}{-4\cos(\theta)\sin(\theta)+\sin(\theta)}$

$\cos^2(\theta)-\sin^2(\theta)$

Correct answer:

$\frac{-2sin^2(\theta)+2\cos^2(\theta)-\cos(\theta)}{-4\cos(\theta)\sin(\theta)+\sin(\theta)}$

Explanation:

The formula for the derivative of a polar function is

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{\frac{\mathrm{dr} }{\mathrm{d} \theta}\sin(\theta)+r\cos(\theta)}{\frac{\mathrm{dr} }{\mathrm{d} \theta}\cos(\theta)-r\sin(\theta)}$

First, we must find the derivative of the function given:

$\frac{\mathrm{dr} }{\mathrm{d} \theta}=-2\sin(\theta)$

Now, we plug in the derivative, as well as the original function, into the above formula to get

$\frac{-2sin^2(\theta)+2\cos^2(\theta)-\cos(\theta)}{-4\cos(\theta)\sin(\theta)+\sin(\theta)}$

Report an Error

Example Question #811 : Calculus Ii

Find the derivative of the following function:

$r=\theta^2+1$

Possible Answers:

$\frac{2\theta\sin(\theta)+(\theta^2+1)(\cos(\theta))}{2\theta\cos(\theta)+(\theta^2+1)(\sin(\theta))}$

$2\theta$

$\frac{2\theta\sin(\theta)+(\theta^2+1)(\cos(\theta))}{2\theta\cos(\theta)-(\theta^2+1)(\sin(\theta))}$

$\frac{2\theta\sin(\theta)-(\theta^2+1)(\cos(\theta))}{2\theta\cos(\theta)+(\theta^2+1)(\sin(\theta))}$

Correct answer:

$\frac{2\theta\sin(\theta)+(\theta^2+1)(\cos(\theta))}{2\theta\cos(\theta)-(\theta^2+1)(\sin(\theta))}$

Explanation:

The derivative of a polar function is given by the following:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}$

First, we must find $\frac{dr}{d\theta}=2\theta$

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Finally, plug in the derivative we just found along with r, the function given, into the above formula:

$\frac{2\theta\sin(\theta)+(\theta^2+1)(\cos(\theta))}{2\theta\cos(\theta)-(\theta^2+1)(\sin(\theta))}$

Report an Error

Example Question #2 : Derivatives Of Polar Form

Find the derivative of the following function:

$r=\cos^2(\theta)$

Possible Answers:

$-2\cos(\theta)\sin(\theta)$

$\frac{-2\cos(\theta)\sin^2(\theta)-\cos^3(\theta)}{-2\cos^2(\theta)\sin(\theta)+\sin(\theta)\cos^2(\theta)}$

$\frac{-2\cos(\theta)\sin(\theta)+\cos^3(\theta)}{-2\cos^2(\theta)\sin(\theta)-\sin(\theta)\cos^2(\theta)}$

$\frac{-2\cos(\theta)\sin^2(\theta)+\cos^3(\theta)}{-2\cos^2(\theta)\sin(\theta)-\sin(\theta)\cos^2(\theta)}$

Correct answer:

$\frac{-2\cos(\theta)\sin^2(\theta)+\cos^3(\theta)}{-2\cos^2(\theta)\sin(\theta)-\sin(\theta)\cos^2(\theta)}$

Explanation:

The derivative of a polar function is given by the following:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}$

First, we must find $\frac{dr}{d\theta}=-2\cos(\theta)\sin(\theta)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x)) \cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Finally, plug in the above derivative and our original function into the above formula:

$\frac{-2\cos(\theta)\sin^2(\theta)+\cos^3(\theta)}{-2\cos^2(\theta)\sin(\theta)-\sin(\theta)\cos^2(\theta)}$

Report an Error

Example Question #2 : Derivatives Of Polar Form

Find the derivative of the following function:

$r=3\theta+e^\theta$

Possible Answers:

$\frac{(3+e^\theta)\sin(\theta)-(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)+(3\theta+e^\theta)\sin(\theta)}$

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)-(3\theta+e^\theta)\sin(\theta)}$

$3+e^\theta$

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)+(3\theta+e^\theta)\sin(\theta)}$

Correct answer:

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)-(3\theta+e^\theta)\sin(\theta)}$

Explanation:

The derivative of a polar function is given by the following:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}$

First, we must find $\frac{dr}{d\theta}=3+e^\theta$

We found the derivative using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Finally, we plug in the above derivative and the original function into the above formula:

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)-(3\theta+e^\theta)\sin(\theta)}$

Report an Error

Example Question #7 : Derivatives Of Polar Form

Find the derivative of the function:

$r=5\cos(2\theta)$

Possible Answers:

$\frac{-10\sin^2(2\theta)+5\cos(2\theta)^2}{-10\sin(2\theta)\cos(\theta)-5\cos(2\theta)\sin(\theta)}$

$\frac{-10\sin(2\theta)\sin(\theta)+5\cos(2\theta)\cos(\theta)}{-10\sin(2\theta)\cos(\theta)+5\cos(2\theta)\sin(\theta)}$

$\frac{-10\sin(2\theta)\sin(\theta)-5\cos(2\theta)\cos(\theta)}{-10\sin(2\theta)\cos(\theta)+5\cos(2\theta)\sin(\theta)}$

$\frac{-10\sin(2\theta)\sin(\theta)+5\cos(2\theta)\cos(\theta)}{-10\sin(2\theta)\cos(\theta)-5\cos(2\theta)\sin(\theta)}$

Correct answer:

$\frac{-10\sin(2\theta)\sin(\theta)+5\cos(2\theta)\cos(\theta)}{-10\sin(2\theta)\cos(\theta)-5\cos(2\theta)\sin(\theta)}$

Explanation:

The derivative of a polar function is given by

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}$

First, we must find the derivative of the function, r:

$\frac{dr}{d\theta}=-10\sin(2\theta)$

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Now, using the derivative we just found and our original function in the above formula, we can write out the derivative of the function in terms of x and y:

$\frac{-10\sin(2\theta)\sin(\theta)+5\cos(2\theta)\cos(\theta)}{-10\sin(2\theta)\cos(\theta)-5\cos(2\theta)\sin(\theta)}$

Report an Error

View Calculus 2 Tutors

Sharif
Certified Tutor

NCCU, Bachelors, Physics and Mathematics.

View Calculus 2 Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View Calculus 2 Tutors

Alejandro
Certified Tutor

University of Western Ontario, Doctorate (PhD), Mathematics.

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Reading Tutors in Phoenix, Spanish Tutors in Denver, Chemistry Tutors in Denver, Algebra Tutors in Phoenix, ACT Tutors in Washington DC, Reading Tutors in New York City, LSAT Tutors in San Francisco-Bay Area, Algebra Tutors in San Francisco-Bay Area, MCAT Tutors in Miami, Computer Science Tutors in Miami

Popular Courses & Classes

SAT Courses & Classes in Phoenix, GRE Courses & Classes in Miami, GMAT Courses & Classes in Boston, SAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Washington DC, MCAT Courses & Classes in Chicago, SSAT Courses & Classes in Denver, GRE Courses & Classes in Phoenix, GRE Courses & Classes in Denver, GMAT Courses & Classes in Phoenix

Popular Test Prep

MCAT Test Prep in Washington DC, GMAT Test Prep in Chicago, MCAT Test Prep in Atlanta, SAT Test Prep in Washington DC, SAT Test Prep in Seattle, ACT Test Prep in Washington DC, SSAT Test Prep in Atlanta, GRE Test Prep in Houston, SSAT Test Prep in Dallas Fort Worth, ISEE Test Prep in New York City

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 2 : Derivatives of Polar Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Derivatives Of Polar Form

Example Question #1 : Derivatives Of Polar Form

Example Question #1 : Derivatives Of Polar Form

Example Question #2 : Derivatives Of Polar Form

Example Question #2 : Derivatives Of Polar Form

Example Question #3 : Derivatives Of Polar Form

Example Question #811 : Calculus Ii

Example Question #2 : Derivatives Of Polar Form

Example Question #2 : Derivatives Of Polar Form

Example Question #7 : Derivatives Of Polar Form

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: